leetcode -- Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / 
            4   8
           /   / 
          11  13  4
         /        
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

public boolean hasPathSum(TreeNode root, int sum) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root == null){
            return false;
        }
        
        if(root.left == null && root.right == null && root.val == sum){
            return true;
        } else {
            return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
        }
    }

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root == null){
            return false;
        }
        ArrayList<ArrayList<Integer>> result = pathSum(root, sum);
        if(result.size() != 0){
            return true;
        }
        return false;
    }
    
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> seqs = new ArrayList<Integer>();
        int curSum = 0;
        
        generatePathSum(root, result, seqs, curSum, sum);
        return result;
    }
    
    public void generatePathSum(TreeNode root, ArrayList<ArrayList<Integer>> result,
        ArrayList<Integer> seqs, int curSum, int expectedSum){
            curSum += root.val;
            seqs.add(root.val);
            
            boolean isLeaf = root.left == null && root.right == null;
            if(curSum == expectedSum && isLeaf){
                ArrayList<Integer> tmp = new ArrayList<Integer>();
                tmp.addAll(seqs);
                result.add(tmp);
            }
            
            if(root.left != null){
                generatePathSum(root.left, result, seqs, curSum, expectedSum);
            }
            
            if(root.right != null){
                generatePathSum(root.right, result, seqs, curSum, expectedSum);
            }
            
            seqs.remove(seqs.size() - 1);
            
        }
}

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null){
            return false;
        }
        return checkPath(root, 0, sum);
    }
    
    public Boolean checkPath(TreeNode root, int curSum, int expectedSum) {
        curSum += root.val;
        boolean isLeaf = root.left == null && root.right == null;
        
        if(curSum == expectedSum && isLeaf){
            return true;
        }
        boolean lResult = false;
        if(root.left != null){
            lResult = checkPath(root.left, curSum, expectedSum);
        }
        boolean rResult = false;
        if(root.right != null){
            rResult = checkPath(root.right, curSum, expectedSum);
        }
        
        if(lResult || rResult){
            return true;
        }
        
        return false;
    }
}